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Since all the <math>f_{n}</math> are increasing, <math> g </math> is increasing and with <math>g(1)</math> is finite, <math>g</math> is finite everywhere, so <math>g(1)=\int_{0}^{1}h(t)dt</math>. | Since all the <math>f_{n}</math> are increasing, <math> g </math> is increasing and with <math>g(1)</math> is finite, <math>g</math> is finite everywhere, so <math>g(1)=\int_{0}^{1}h(t)dt</math>. | ||
| − | Conclusion, <math> | + | Conclusion, <math>g</math> is AC. |
Revision as of 10:36, 10 July 2008
Since all the $ f_{n} $ are AC, there exists $ f_{n}^{'} $ such that $ f_{n}(x)=f_{n}(x)-f_{n}(0)=\int_{0}^{x}f_{n}^{'}(t)dt $ and $ f_{n}^{'} $ are nonnegative almost everywhere.
Let $ h(t)=\sum_{k=1}^{what the hell is this}f_{k}^{'}(t) $
Let $ g_{n}(x)= \sum_{k=1}^{n}f_{k}(x)=\sum_{1}^{n}\int_{0}^{x}f_{k}^{'}(t)dt=\int_{0}^{x}(\sum_{k=0}^{n}f_{k}^{'}(t))dt $
Lebesgue Monotone Convergence Theorem gives us $ g(x)=\int_{0}^{x}h(t)dt $
Since all the $ f_{n} $ are increasing, $ g $ is increasing and with $ g(1) $ is finite, $ g $ is finite everywhere, so $ g(1)=\int_{0}^{1}h(t)dt $. Conclusion, $ g $ is AC.
